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Let y be a transient state. Show that for all x
$$\sum_{n=0}^{\infty}P^n(x,y) \leq \sum_{n=0}^{\infty}P^n(y,y)$$

Hint: Use $\sum_{n=1}^{\infty}P^n(x,y)=\frac{\rho_{xy}}{1-\rho_{yy}}$

I tried to sum one and use the property $0 \leq \rho_{xy} \leq 1$ but I could not conclude. Any ideas?

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  • $\begingroup$ Doesn't $\sum_{i=1}^\infty P^n(x,y)$ obviously diverge? $\endgroup$ – Kenta S Mar 25 at 20:20
  • $\begingroup$ mmm why? y is transient so there is no problem, right? $\endgroup$ – Bayesian guy Mar 25 at 21:09
  • $\begingroup$ this doesn't make any sense $\sum_{i=0}^{\infty}P^n(x,y) \leq \sum_{i=0}^{\infty}P^n(y,y)$ the indexing in the sum is over i, but $i$ doesn't show up in there, only $n$ and some y, evidently neither of which are related to $i$ $\endgroup$ – user8675309 Mar 25 at 23:50
  • $\begingroup$ my bad, it has to be n instead of i $\endgroup$ – Bayesian guy Mar 26 at 4:33
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I did it... We know that $$P^0(x,y)=P[X_0=y|X_0=x]=\begin{cases} 1 & if x=y\\ 0 & if x \neq y \end{cases}$$ In other hand , let's see that \begin{align*} \sum_{n=0}^{\infty}P^n(y,y)&=P^0(y,y)+\sum_{n=1}^{\infty}P^n(y,y)\\ &=1+\sum_{n=1}^{\infty}P^n(y,y)\\ &=1+\frac{\rho_{yy}}{1-\rho_{yy}} \tag{Hint}\\ &=\frac{1}{1-\rho_{yy}}. \end{align*}

Therefore, \begin{align*} \sum_{n=0}^{\infty}P^n(x,y)&=P^0(x,y)+\sum_{n=1}^{\infty}P^n(x,y)\\ &=0+\sum_{n=1}^{\infty}P^n(x,y)\\ &=\frac{\rho_{xy}}{1-\rho_{yy}} \tag{Hint}\\ &\leq \frac{1}{1-\rho_{yy}}. \tag{$0 \leq \rho_{xy} \leq 1$}\\ &=\sum_{n=0}^{\infty}P^n(y,y). \end{align*}

Hence, $$\sum_{n=0}^{\infty}P^n(x,y) \leq \sum_{n=0}^{\infty}P^n(y,y).$$ Q.E.D.

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